VJC 2023 H2 9744 P2 Ans

1 Victoria Junior College Biology Department 2023 Prelims H2 Paper 2 – Proposed Answers Marking abbreviations: A: Accept, R: Reject, AW: alternative wording, AVP: Any valid point, NAQ: not answering question, ECF: error carried forward 1 The cell surface membrane regulates the movement of different substances via different mechanisms. Table 1.1 shows the results from an investigation to determine how two different drugs A and B enter animal cells. Table 1.1 concentration of drug A in container / arbitrary units concentration of drug A inside the cells after 5 minutes / arbitrary units concentration of drug B in container / arbitrary units concentration of drug B inside the cells after 5 minutes / arbitrary units 0 0 0 0 5 5 5 4 10 10 10 7 15 15 15 11 20 19 20 13 25 24 25 13 Cells which did not contain either drug were placed into separate containers. Different concentrations of each drug were added and after 5 minutes, the drug concentration in the cells was measured. (a) Based on the data shown in Table 1.1, (i) state the chemical nature of drugs A and B [1] • A is lipid soluble/hydrophobic whereas B is polar/hydrophilic; (ii) explain the difference between the transport of drugs A and B. [3] • Data to support differences: Concentration of the drug A inside the cells matches/similar to the external concentration while for drug B, as external concentration increases, the internal concentration also increases but reaches a plateau; • Drug A enters by simple diffusion whereas B enters via facilitated diffusion; OR Drug A diffuses across the phospholipid bilayer/ hydrophobic core of cell membrane into cell, but B requires carrier/ channel proteins to enter cell; • A: moves down concentration gradient from outside of the cell into the cell until equilibrium is reached; • B: (Idea of) the number of carrier/ channel proteins on cell membrane are fixed/ saturated;

2 (b) Viruses make use of certain properties of the cell membrane to facilitate its own release from the host cell. Fig. 1.1 shows such a process. (i) Explain clearly how the structure of the cell membrane facilitates this process. [3] • Phospholipid bilayer structure of the cell membrane provides fluidity / lateral movement within their own monolayer; • P hospholipids held by weak hydrophobic interactions between the hydrocarbon tails of the phospholipids; • allowing the cell surface membrane to change its shape / be pushed outwards – budding; • (idea of) allowing viral glycoproteins to be embedded in cell surface membrane as exit points; (ii) Contrast the process seen in Fig 1.1 with the process in which HIV virus enters the host cell. [2] Process shown in Fig 1.1 Process which HIV virus enters the host cells [Any 2 of the following] • Budding involving evagination of host cell surface membrane • Inv olves removal / loss of cell membrane; • Involves viral glycoproteins on cell surface membrane as exit points for budding • Fusion of viral envelope with host cell surface membrane; • A ddition/ gain of cell membrane; • gp120 on viral envelope binding to CD4 receptor on cell surface membrane; R: Budding vs no budding, fusion vs no fusion, loss vs no loss of membrane, etc. – when there is an equivalent process, students should describe the difference in that feature.

3 2 (a) Starch molecules are the main storage molecules in many types of cereal grain, such as the grain of the barley plant. When the seed inside a barley grain germinates, enzymes are synthesised to catalyse the hydrolysis of the starch molecules. (i) Starch is a mixture of two different molecules. Name these two molecules. [1] • Amylose and amylopectin; (ii) Describe two features that allow the starch to be a good storage molecule. [2] Any 2: • Can be coiled and folded into compact shape, maximising storage space/ AW; • Can be easily hydrolysed to ( α-)glucose (R: energy) when they are required by enzymes/ amylase breaking glycosidic bonds/ to produce large amount of energy in respiration; • Has large size which makes it insoluble in water so will not affect the water potential of cell; • AVP (b) In the space below, draw the molecular structure of maltose and show how it can be catalysed by maltase to produce glucose molecules. [3] Diagram includes the following: • Maltose drawn correctly with 1,4 glycosidic bond labelled; • Water molecule being added and/ or name the process as hydrolysis; • Products drawn correctly, two alpha glucose; When producing sugar syrups, there are advantages in using enzymes extracted from microorganisms. For example, some enzymes extracted from microorganisms are heat stable. Heat-stable enzymes are used to increase productivity because the reactions can be carried out at higher temperatures. (c) Suggest one other advantage of using enzymes obtained from microorganisms, rather than enzymes extracted from barley seeds, in the production of sugar syrups. [1] Any one valid suggestion; • e.g. easier to extract • idea that microorganisms can be cultured in large quantities and produce large amounts of enzyme • higher rate of reaction • active over a greater temperature range

4 (d) Fig. 2.2 is a graph showing how the activity of α- amylase extracted from barley seeds changes as the temperature increases from 10 °C to 66 °C. (i) Explain the effect of temperature on the activity of α-amylase extracted from barley seeds, as shown in Fig. 2.2. [3] Any three from: 1. increase in temperature from 10 to 48 °C, increases kinetic energy of, molecules / substrate and enzyme ; A moving around faster 2. (increase in temperature) increases rate of, successful/ effective collisions between enzyme and substrate / enzyme– substrate complex formation / substrate binding to active site ; OR at 48 °C (47 or 49) ref. to, optimum temperature / maximum enzyme activity / active sites binding substrate at maximum capacity / AW ; 3. (at temperatures higher than optimum) ref. to denaturation / loss of shape of active site (and decrease in activity) ; 4. AVP ; e.g. weaker / hydrogen/ ionic, bonds break (and ref. to denaturation) R disulfide bond – covalent bond, strong not easily broken by high temperature (ii) Sketch on Fig. 2.2 the curve that would be obtained using the heat stable α -amylase enzyme extracted from microorganisms. [2] • optimum / peak / max activity / at 100% and to right of original ; • line beyond optimum must be to right and, reach vertical axis at 50% or below / reach horizontal axis ; Example:

5 3 Collagen is present in large quantities in connective tissue and provides tendons and ligaments with tensile strength. Fig. 3.1 below shows how tropocollagen molecules is organised in collagen. Fig. 3.1 (a) Identify structure X and explain how it allows collagen to serve a structural role. [3] 1. X is collagen fibril 2. Staggered arrangement of tropocollagen within the collagen fibril eliminates areas of weakness; 3. Covalent cross-links between the tropocollagen to strengthen the structure, contributing to high tensile strength; 4. Aggregation of collagen fibrils into collagen fibres, contribute to high tensile strength (b) Each of the three polypeptide in the tropocollagen is a kinked helix whereas in another macromolecule, cellulose, each chain is a straight chain. (i) Explain how the kinked helix in tropocollagen and the straight chain in cellulose contribute to high tensile strength. [3] 1. Both kinked helix and cellulose are strand-like structures which provide high/more surface area for crosslinking/ hydrogen bonds formation. 2. Kinked helix allows the three polypeptide to wind more tightly around each other and thus require more force to break them. 3. Straight chain in cellulose allows most hydroxyl groups of the glucose residue